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Maths-

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Question

The solution of the differential equation $fraction numerator d y over denominator d x end fraction plus y equals cos invisible function application x$ is

$y = \frac{1}{2} (\cos x + \sin x) + c e^{- x}$
$y = \frac{1}{2} (\cos x - \sin x) + c e^{- x}$
$y = \cos x + \sin x + c e^{- x}$
None of these

The correct answer is: $y equals fraction numerator 1 over denominator 2 end fraction left parenthesis cos invisible function application x plus sin invisible function application x right parenthesis plus c e to the power of negative x end exponent$

It is linear equation of the form $fraction numerator d y over denominator d x end fraction plus P y equals Q$
So, I.F. $equals e to the power of not stretchy integral 1 d x end exponent equals e to the power of x end exponent$
Hence solution is $y. e to the power of x end exponent equals not stretchy integral cos invisible function application x. e to the power of x end exponent d x plus c$
Þ $y equals fraction numerator 1 over denominator 2 end fraction left parenthesis cos invisible function application x plus sin invisible function application x right parenthesis plus c e to the power of negative x end exponent$ .

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